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+\documentclass[a4paper,french,12pt]{article}
+
+\title{Logique Programmable --- Exercices}
+\author{Catherine MARECHAL --- \href{mailto:catherine.marechal@efrei.fr}{\nolinkurl{catherine.marechal@efrei.fr}}}
+\date{Dernière compilation~: \today{} à \currenttime}
+
+\usepackage{../../cours}
+
+\begin{document}
+
+\maketitle
+
+\section{Simplification des fonctions logiques}
+    \subsection{Exercice 1}
+    \subsection{Exercice 2}
+    \subsection{Exercice 3}
+    \subsection{Exercice 4}
+    \subsection{Exercice 5}
+    \subsection{Exercice 6}
+    \subsection{Exercice 7}
+
+\section{Circuits de logique combinatoire}
+    \subsection{Exercice 1}
+
+        \includegraphics[width=0.6\linewidth]{./img/2.1.png}
+
+    \subsection{Exercice 2}
+
+        a/ Élaborer l'équation logique $F(S_1,S_0,C,a,b)$ du circuit ci-dessous.
+
+        \includegraphics[width=\linewidth]{./img/2.2.png}
+
+        \begin{equation*}
+            F = \bar{c} + \overline{S_0}(ab + \bar{a}\bar{b}S_1)
+        \end{equation*}
+
+        b/ Compléter le tableau suivant pour C = 0~:
+
+        $F = \bar{c}$
+
+        \begin{tabularx}{0.8\linewidth}{cccY}
+            \toprule
+            C & $S_1$ & $S_0$ & $F(a,b)$ \\
+            \midrule
+            0 & 0 & 0 & 1 \\
+            0 & 0 & 1 & 1 \\
+            0 & 1 & 0 & 1 \\
+            0 & 1 & 1 & 1 \\
+            \bottomrule
+        \end{tabularx}
+
+        c/ Compléter le tableau suivant pour C = 1~:
+
+        $F = \overline{S_0}(ab + \bar{a}\bar{b}S_1)$
+
+        \begin{tabularx}{0.8\linewidth}{cccY}
+            \toprule
+            C & $S_1$ & $S_0$ & $F(a,b)$ \\
+            \midrule
+            1 & 0 & 0 & $ab$ \\
+            1 & 0 & 1 & 0 \\
+            1 & 1 & 0 & $ab + \bar{a}\bar{b}$ \\
+            1 & 1 & 1 & 0 \\
+            \bottomrule
+        \end{tabularx}
+
+    \subsection{Exercice 3}
+
+        Soit le schéma de $H(A,B,C,D)$ utilisant un multiplexeur à 3 entrées d'adresse.
+
+        \includegraphics[width=0.6\linewidth]{./img/2.3.png}
+
+        a/ Donner la table de vérité de la fonction $H$.
+
+        \begin{tabularx}{0.7\linewidth}{X|Y|YYY|Y}
+            \toprule
+                   & D & C & B & A & H \\
+            \midrule
+            0      & 0 & 0 & 0 & 0 & 0 \\
+            1      & 0 & 0 & 0 & 1 & 0 \\
+            2      & 0 & 0 & 1 & 0 & 0 \\
+            3      & 0 & 0 & 1 & 1 & 0 \\
+            4      & 0 & 1 & 0 & 0 & 0 \\
+            5      & 0 & 1 & 0 & 1 & 0 \\
+            6      & 0 & 1 & 1 & 0 & 1 \\
+            7      & 0 & 1 & 1 & 1 & 0 \\
+            \midrule
+            8 (0)  & 1 & 0 & 0 & 0 & 0 \\
+            9 (1)  & 1 & 0 & 0 & 1 & 0 \\
+            10 (2) & 1 & 0 & 1 & 0 & 1 \\
+            11 (3) & 1 & 0 & 1 & 1 & 1 \\
+            12 (4) & 1 & 1 & 0 & 0 & 0 \\
+            13 (5) & 1 & 1 & 0 & 1 & 1 \\
+            14 (6) & 1 & 1 & 1 & 0 & 1 \\
+            15 (7) & 1 & 1 & 1 & 1 & 0 \\
+            \bottomrule
+        \end{tabularx}
+
+        b/ Exprimer $H(A,B,C,D)$ sous la forme disjonctive.
+
+        \begin{equation*}
+            H(A,B,C,D) = \bar{A}BC\bar{D} + \bar{A}B\bar{C}D + AB\bar{C}D + A\bar{B}CD + \bar{A}BCD
+        \end{equation*}
+
+    \subsection{Exercice 4}
+
+\end{document}